Calculus

A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45' and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the intitial speed of the ball, and how high does it rise?

So far i have got a few formulas, but i am still at a loss at where to start and when i start to put my formulas together i start to get zeros because everything is canceling out.

For a ball caught at the same altitude that it is hit, the horizontal distance travelled is
X = (Vo^2/g)sin(2A)
where A is the "launch" angle from horizontal.
In your case 2A = 45 degrees, so
Vo^2 = g X

Solve for Vo.

The height it rises is

H = (Vo sinA)^2/g = Vo^2/(2g) = X/2

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